Question: The equation of a circle $C$ is $x^2+y^2+14x-14y+62 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2+14x) + (y^2-14y) = -62$ $(x^2+14x+49) + (y^2-14y+49) = -62 + 49 + 49$ $(x+7)^{2} + (y-7)^{2} = 36 = 6^2$ Thus, $(h, k) = (-7, 7)$ and $r = 6$.